Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
int main(){
int N;
cin>>N;
int p=N+1;
double  P[p];
for(int i=1;i<=N;i++){cin>>P[i];}

double dp[N+1][N+1];
for(int i=0;i<=N;i++)dp[i][0]=0;
dp[0][0]=1;
for(int i=1;i<=N;i++)dp[0][i]=dp[0][i-1]*(1-P[i]);

for(int i=1;i<=N;i++)for(int j=1;j<=N;j++){
dp[i][j]=dp[i-1][j-1]*(P[j])+dp[i][j-1]*(1-P[j]);
}

long double out=0;
for(int i=((N/2)+1);i<=N;i++)
{
out+=dp[i][N];
}
printf("%.9Lf",out);
}